A network of resistors is connected to a 12V battery as shown in fig.

a) Calculate the equivalent resistance of the network.
b) Obtain current in `12Omega` and `6Omega` resistors.

Explanation

`R_(1)=12Omega`
`R_(2)=6Omega`
`V=12V`
`R_(P)=?`
`I_(1)=?`
`I_(2)=?`
The equivalent resistance of the parallel combination is
`(1)/(R_(p))=(1)/(R_(1))+(1)/(R_(2))`
`R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))=(12xx6)/(12+6)`
`R_(p)=4Omega`
The current through resistor `R_(1)=12Omega` is
`V=I_(1)R_(1)`
`I_(1)=(V)/(R_(1))=(12)/(12)=1A`
The current through resistor `R_(2)=6Omega` is
`V=I_(2)R_(2)`
`I_(2)=(V)/(R_(2))=(12)/(6)=2A`